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FOLDL - Lists - Haskell
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Haskell - Lists - Foldl
foldl

This is how the left fold is implemented. Notice how the order of the arguments in the step function is flipped compared to foldr (the right fold):

foldl :: (b -> a -> b) -> b -> [a] -> b
foldl f acc []     =  acc
foldl f acc (x:xs) =  foldl f (f acc x) xs         -- = foldl f (acc `f` x) xs

The left fold, foldl, associates to the left. That is:

foldl (+) 0 [1, 2, 3]     -- is equivalent to ((0 + 1) + 2) + 3

The reason is that foldl is evaluated like this (look at foldl's inductive step):

foldl (+) 0 [1, 2, 3]                        --  foldl (+)    0   [ 1,   2,   3 ]
foldl (+) ((+) 0 1) [2, 3]                   --  foldl (+)   (0 + 1)   [ 2,   3 ]
foldl (+) ((+) ((+) 0 1) 2) [3]              --  foldl (+)  ((0 + 1) + 2)   [ 3 ]
foldl (+) ((+) ((+) ((+) 0 1) 2) 3) []       --  foldl (+) (((0 + 1) + 2) + 3) []
((+) ((+) ((+) 0 1) 2) 3)                    --            (((0 + 1) + 2) + 3)

The last line is equivalent to ((0 + 1) + 2) + 3. This is because (f a b) is the same as (afb) in general, and so ((+) 0 1) is the same as (0 + 1) in particular.

 
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